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NEET CHEMISTRYMedium

If at a given instant, for the reaction 2N2O5→4NO2+O22N_2O_5 \rightarrow 4NO_2 + O_2 rate and rate constant are 1.02×10−41.02 \times 10^{-4} and 3.4×10−5 sec−13.4 \times 10^{-5} \text{ sec}^{-1} respectively, then the concentration of N2O5N_2O_5 at that time will be:

A

1.732 mol L−11.732 \text{ mol L}^{-1}

B

3.0 mol L−13.0 \text{ mol L}^{-1}

C

1.02×10−4 mol L−11.02 \times 10^{-4} \text{ mol L}^{-1}

D

3.4×105 mol L−13.4 \times 10^5 \text{ mol L}^{-1}

Step-by-Step Solution

The unit of the rate constant kk is given as sec−1\text{sec}^{-1}, which indicates that the decomposition of N2O5N_2O_5 is a first-order reaction.

For a first-order reaction, the rate law is expressed as: Rate=k[N2O5]\text{Rate} = k[N_2O_5]

Given: Rate=1.02×10−4 mol L−1 sec−1\text{Rate} = 1.02 \times 10^{-4} \text{ mol L}^{-1} \text{ sec}^{-1} Rate constant, k=3.4×10−5 sec−1k = 3.4 \times 10^{-5} \text{ sec}^{-1}

Rearranging the rate law to solve for the concentration of N2O5N_2O_5: [N2O5]=Ratek[N_2O_5] = \frac{\text{Rate}}{k} [N2O5]=1.02×10−43.4×10−5=10.2×10−53.4×10−5=3.0 mol L−1[N_2O_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} = \frac{10.2 \times 10^{-5}}{3.4 \times 10^{-5}} = 3.0 \text{ mol L}^{-1}

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